Remark 2

Suppose that $H$ is generated by $H_0$ and $H_1$; then for every $n$, one has $H_{n+1} = H_1.H_n$, hence $H_{\geq n+1} = H_1.H_{\geq n}$ and $S^{-1}H_{\geq n+1} = H_1.S^{-1}H_{\geq n}$. It follows that the filtration $(S^{-1}H_{\geq n})$ of $S^{-1}H$ is the $S^{-1}H_{\geq 1}$-adic filtration.